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More Conservation of Momentum Contents: Whiteboards:

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Presentation on theme: "More Conservation of Momentum Contents: Whiteboards:"— Presentation transcript:

1 More Conservation of Momentum Contents: Whiteboards:

2 A 3452 kg truck going 23.0 m/s collides with a 1253 kg car at rest. What is the velocity of the wreckage just after the collision? ( 3452kg )( 23.0m/s)=(3452kg+1253kg)v 79396kgm/s = (4705kg)v (79396kgm/s)/(4705kg) = 16.9 m/s 16.9 m/s

3 Football Player A has a mass of 89.0 kg and is running West at 7.60 m/s. He collides with Football Player B (92.0 kg) and the wreckage has a velocity of 1.50 m/s to the West. What was the velocity of B before the collision? (Speed and direction) 4.40 m/s to the East, or -4.40 m/s to the West

4 86.0 kg Thor is standing on a 35.0 kg cart, and is holding a 13.0 kg hammer. Everything is moving to the right at 1.60 m/s. After he throws the hammer, he and his cart are moving 2.40 m/s to the right. What speed and in what direction did Thor throw the hammer? 5.85 m/s to the left

5 A 215 gram bullet going 542 m/s goes through a stationary 3.40 kg block of wood. After this, the block is traveling at 12.0 m/s in the same direction, (a) what is the bullet’s velocity? (. 215kg)(542 m/s) = (3.4kg)(12m/s) + (.215kg)v 116.53kgm/s = 40.8kgm/s + (.215kg)v 75.73kgm/s = (.215kg)v (75.73kgm/s)/(.215kg) = 352 m/s 352 m/s

6 A 215 gram bullet going 542 m/s goes through a stationary 3.40 kg block of wood. After this, the block is traveling at 12 m/s in the same direction, (b) if ripping through the block took.0250 s, what force did the bullet exert on the block? use m  v = F  t: (3.4 kg)(12m/s) = (F)(.025 s) F = (3.4 kg)(12m/s)/(.025 s) = 1632 N = 1630 N 1630 N

7 A 2450. kg airplane (including ammunition) going 52.40 m/s fires a 5.120 kg shell forward at 890.0 m/s. What is the velocity of the plane afterward? (2450kg)(52.40m/s) = (2444.88 kg)v+(5.12kg)890m/s 128,380kgm/s = (2444.88 kg)v +4556.8kgm/s 123,823.2kgm/s = (2444.88 kg)v (123,823.2kgm/s)/( 2444.88 kg ) = 50.65 m/s 50.65 m/s

8 A 153 gram bullet going 452 m/s goes through the first of two stationary 3.50 kg blocks of wood, and sticks in the second. After this, the first block is traveling at 6.50 m/s in the same direction, what speed are the second block and bullet going? (.153kg)(452m/s) = (3.5kg)(6.5m/s)+(.153kg+3.5kg)v 69.156kgm/s = 22.75kgm/s + (3.653kg)v 47.406kgm/s = (3.653kg)v (47.406kgm/s)/(3.653kg) = 12.7 m/s 12.7 m/s

9 A 153 gram bullet going 452 m/s goes through the first of two stationary 3.50 kg blocks of wood, and sticks in the second. After this, the first block is traveling at 6.50 m/s in the same direction, What is the bullet’s velocity between the blocks? (.153kg)(452m/s) = (3.5kg)(6.5m/s)+(.153kg)v 69.156kgm/s = 22.75kgm/s + (.153kg )v 46.406kgm/s = (.153kg )v (46.406kgm/s)/(.153kg ) = 303.31 = 303 m/s 303 m/s

10 A 153 gram bullet going 452 m/s goes through two stationary 3.50 kg blocks of wood. After this, the first block is traveling at 6.50 m/s and the second is going 8.30 m/s in the same direction. What is the bullet’s velocity after this? (.153kg)(452m/s)= (3.5kg)(6.5m/s)+(3.5kg)(8.3m/s)+(.153kg)v 69.156kgm/s = 22.75kgm/s+29.05kgm/s+(.153kg )v 17.356kgm/s = (.153kg )v (17.356kgm/s)/(.153kg ) = 113.44 = 113 m/s 113 m/s

11 A 153 gram bullet going 452 m/s is heading toward two stationary 3.50 kg blocks of wood. What is the velocity if the bullet sticks in the first block, and the two blocks then slide and stick together? (.153kg)(452m/s)= (3.5kg+3.5kg+.153kg)v 69.156kgm/s = (7.153kg)v (69.156kgm/s)/(.153kg ) = 9.668 = 9.67 m/s 9.67 m/s

12 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 1. What is the velocity of car A when he is in midair? 0 = (60kg)(+5.2 m/s) + (560 kg)v v = -.557 m/s -.56 m/s A B

13 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 2. What is the velocity of car B when he leaves it? (60 kg)(5.2 m/s) + 0 = (60 kg)(3.4 m/s) + (780 kg)v v = +.138 m/s +.14 m/s A B

14 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 3. What would have been the velocity of car B had he remained there, and not jumped off? +.37 m/s A B (60 kg)(5.2 m/s) + 0 = (60 kg+ 780 kg)v v = +.371 m/s

15 60. kg Brennen is playing on two flatbed rail cars initially at rest. Car A has a mass of 560 kg and B 780 kg. He reaches a velocity of +5.2 m/s on A, before jumping to B where he slows to +3.4 m/s before jumping off the other end. The cars are uncoupled, and rest on a frictionless track 4. What would the velocity of car B have been had he jumped off the back of it to give himself a velocity of zero? (60 kg)(5.2 m/s) + 0 = (60 kg)(0 m/s)+ (780 kg)v v = +.40 m/s +.40 m/s A B


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